Question

There are four numbers such that average of first two numbers is 1 more than the first number, average of first three numbers is 2 more than average of first two numbers, and average of first four numbers is 3 more than average of first three numbers. Then, the difference between the largest and the smallest numbers, is

Answer 1

Correct Answer: 15

Let the four numbers be $a, b, c, d$.

Step 1. The average of the first two numbers is $\frac{a+b}{2}$, and it is 1 more than $a$, so:
$\frac{a+b}{2} = a+1 \implies a+b = 2a+2 \implies b = a+2$.

Step 2. The average of the first three numbers is $\frac{a+b+c}{3}$, and it is 2 more than the average of the first two, so:
$\frac{a+b+c}{3} = \frac{a+b}{2} + 2 \implies \frac{a+b+c}{3} = a+1+2 = a+3$.
$a+b+c = 3(a+3) = 3a+9 \implies c = 3a+9 - (a+b) = 3a+9 - (a+a+2) = 2a+7$.

Step 3. The average of the first four numbers is $\frac{a+b+c+d}{4}$, and it is 3 more than the average of the first three numbers, so:
$\frac{a+b+c+d}{4} = \frac{a+b+c}{3} + 3 \implies \frac{a+b+c+d}{4} = a+3+3 = a+6$.
\(a+b+c+d = 4(a+6) = 4a+24 \)
\(\implies\) \(d = 4a+24 - (a+b+c) = 4a+24 - (a+a+2+2a+7) = 15.\)

The numbers are $a, a+2, 2a+7, 15$. The largest number is 15, and the smallest is $a$.
Thus, the difference is:
$15 - a = 15$.